- #1

Lebnm

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Is it correct? To me, it was necessary to proof also that ##g(g g^{-1}) = g##, because with this I could use the fact that the identity of a group is unique.

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- Thread starter Lebnm
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- #1

Lebnm

- 31

- 1

Is it correct? To me, it was necessary to proof also that ##g(g g^{-1}) = g##, because with this I could use the fact that the identity of a group is unique.

- #2

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Another way to define a group is by demanding that the equations ##ax=b## and ##xa=b## can uniquely be solved (IIRC).

- #3

WWGD

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